What is area y=2x^2,y=16x-32,y=0 ?

The answer to area y=2x^2,y=16x-32,y=0 is 32/3

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area y=2x^2,y=16x-32,y=0

area

y = 2 x^{2}, y = 16 x - 32, y = 0

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

2 x^{2} = 16 x - 32 : x = 4

2 x^{2} = 0 : x = 0

16 x - 32 = 0 : x = 2

= \int_{0}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{2}^{4} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{0}^{2} 2 x^{2} - 0 d x + \int_{2}^{4} 16 x - 32 - 2 x^{2} d x

\int_{0}^{2} 2 x^{2} - 0 d x = \frac{16}{3}

\int_{2}^{4} 16 x - 32 - 2 x^{2} d x = \frac{16}{3}

= \frac{16}{3} + \frac{16}{3}

Simplify

= \frac{32}{3}

\frac{d y}{d t} = \frac{y + 1}{t + 1}

\frac{\partial}{\partial x} (6 - arctan (7 x^{2} + 2 y^{2}))

\int t ln (t + 2) d t

\frac{d}{d t} (sin (5 t))

y^{^{'}} = 2 x - 3 y

What is area y=2x^2,y=16x-32,y=0 ?
The answer to area y=2x^2,y=16x-32,y=0 is 32/3